1) First is an odd function because f(-x)=-f(x)
3) 3d is an even function because f(-x)=f(x)
2) 2d looks like sin function and it is also odd.
Answer: x = 8
(the correct answer was not provided as an option)
<u>Step-by-step explanation:</u>
![log_2(x-6)+log_2(x-4)=log_2(x)\\\\log_2[(x-6)(x-4)]=log_2(x)\\\\(x-6)(x-4)=x\\\\x^2-10x+24=x\\\\x^2-11x+24=0\\\\(x-3)(x-8)=0\\\\x=3\qquad x=8\\\\\\\text{The term after the log symbol (inside the parenthesis) must be greater than 0!}\\\\Check:\\3-6>0\ \text{FALSE --- so 3 is not a valid solution}\\\\8-6>0\ \text{TRUE}\\8-4>0\ \text{TRUE}\\8>0\ \text{TRUE --- 8 is a valid solution because it is true for all}](https://tex.z-dn.net/?f=log_2%28x-6%29%2Blog_2%28x-4%29%3Dlog_2%28x%29%5C%5C%5C%5Clog_2%5B%28x-6%29%28x-4%29%5D%3Dlog_2%28x%29%5C%5C%5C%5C%28x-6%29%28x-4%29%3Dx%5C%5C%5C%5Cx%5E2-10x%2B24%3Dx%5C%5C%5C%5Cx%5E2-11x%2B24%3D0%5C%5C%5C%5C%28x-3%29%28x-8%29%3D0%5C%5C%5C%5Cx%3D3%5Cqquad%20x%3D8%5C%5C%5C%5C%5C%5C%5Ctext%7BThe%20term%20after%20the%20log%20symbol%20%20%28inside%20the%20parenthesis%29%20must%20be%20greater%20than%200%21%7D%5C%5C%5C%5CCheck%3A%5C%5C3-6%3E0%5C%20%5Ctext%7BFALSE%20---%20so%203%20is%20not%20a%20valid%20solution%7D%5C%5C%5C%5C8-6%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8-4%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8%3E0%5C%20%5Ctext%7BTRUE%20---%208%20is%20a%20valid%20solution%20because%20it%20is%20true%20for%20all%7D)
Answer: The angle is in the second quandrant
Step-by-step explanation:
If sin of theta is greater than 0, the angle theta has to be in either the first or second quandrant. If tangent of theta is less than 0, it cannot be in the first or third quandrant. Therefore, we know the angle is in the second quandrant.
